To find the value of the limit $$L$$, we first determine the derivative function $$f'(x)$$ using the first principles of differentiation and the given functional equation:
1. Find the general expression for the derivative $$f'(x)$$:
By the definition of the derivative, we have:
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$
Using the given functional equation to substitute for $$f(x + h)$$, we get:
$$f'(x) = \lim_{h \to 0} \frac{f(x)e^{h} + f(h)e^{x} + 4xh - f(x)}{h}$$
Rearrange and group the terms to isolate the limits:
$$f'(x) = \lim_{h \to 0} \frac{f(x)(e^{h} - 1) + f(h)e^{x} + 4xh}{h}$$
$$f'(x) = f(x) \cdot \lim_{h \to 0} \left(\frac{e^{h} - 1}{h}\right) + e^{x} \cdot \lim_{h \to 0} \left(\frac{f(h) - f(0)}{h}\right) + 4x$$
Since $$\lim_{h \to 0} \frac{e^h - 1}{h} = 1$$ and $$\lim_{h \to 0} \frac{f(h) - f(0)}{h} = f'(0) = 2$$, substituting these standard limits yields the first-order differential equation:
$$f'(x) = f(x) + 2e^{x} + 4x$$
2. Find the second derivative function $$f''(x)$$:
Differentiate both sides of our newly found expression with respect to $$x$$:
$$f''(x) = f'(x) + 2e^{x} + 4$$
3. Evaluate the given limit expression $$L$$:
We are asked to evaluate the limit:
$$L = \lim_{x \to 1} \frac{f'(x) - f'(1)}{x^2 - 1}$$
We can factor the denominator as $$(x - 1)(x + 1)$$ and rewrite the expression to reveal a standard derivative structure:
$$L = \lim_{x \to 1} \left[ \frac{f'(x) - f'(1)}{x - 1} \cdot \frac{1}{x + 1} \right]$$
By the definition of the second derivative, the first part of the limit is exactly $$f''(1)$$:
$$L = f''(1) \cdot \frac{1}{1 + 1} = \frac{f''(1)}{2}$$
4. Compute the value of $$f''(1)$$:
Using the differential equations established in the earlier steps, we substitute $$x = 1$$:
$$f''(1) = f'(1) + 2e^{1} + 4$$
Now substitute the expression for $$f'(1)$$ into the equation above:
$$f'(1) = f(1) + 2e^{1} + 4(1)$$
$$f''(1) = (f(1) + 2e + 4) + 2e + 4 = f(1) + 4e + 8$$
To find $$f(1)$$, we solve the linear differential equation $$f'(x) - f(x) = 2e^x + 4x$$. The integrating factor is $$e^{-x}$$.
Multiplying both sides by the integrating factor gives:
$$\frac{d}{dx}[f(x)e^{-x}] = 2 + 4xe^{-x}$$
Integrating both sides from 0 to 1:
$$\left[f(x)e^{-x}\right]_0^1 = \int_0^1 (2 + 4xe^{-x}) dx$$
$$f(1)e^{-1} - f(0)e^0 = [2x]_0^1 + 4 \cdot \left([-xe^{-x}]_0^1 + \int_0^1 e^{-x} dx\right)$$
$$f(1)e^{-1} = 2 + 4 \cdot \left(-e^{-1} + [-e^{-x}]_0^1\right)$$
$$f(1)e^{-1} = 2 + 4 \cdot (-e^{-1} - e^{-1} + 1)$$
$$f(1)e^{-1} = 2 - 8e^{-1} + 4 = 6 - 8e^{-1}$$
Multiplying the entire equation by $$e$$ gives:
$$f(1) = 6e - 8$$
Now, substitute $$f(1)$$ back into the expression for $$f''(1)$$:
$$f''(1) = (6e - 8) + 4e + 8 = 10e$$
$$ L = 5e $$
$$ \frac{L}{e} = 5$$
91- 6303239042
